# Thread: limit definition of derivative

1. ## limit definition of derivative

3)Use the limit definition of derivative and find, $\displaystyle y'$
for $\displaystyle \frac{1}{x+1}$

I know have the answer because we are taught the short cut method in school where we multiply the coefficient by the power and decrease by one. But I want to see how this is done the longer method.

2. Find this limit.
$\displaystyle \displaystyle\lim _{h \to 0} \frac{{\frac{1} {{x + 1 + h}} - \frac{1} {{x + 1}}}} {h}$

3. I end up with:

$\displaystyle \frac{h^2}{(x+1)^2 +h(x+1)}$

4. Originally Posted by elieh
I end up with:$\displaystyle \frac{h^2}{(x+1)^2 +h(x+1)}$
You made some algebra mistakes.
$\displaystyle \dfrac{\frac{1}{x+h+1}-\frac{1}{x+1}}{h}=\dfrac{\frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}}{h}$

5. $\displaystyle \frac{-h^2}{(x+1)^2 + h(x+1)}$

If h is on top and it tends to 0 then the limit is 0...

6. The answer is not zero.
Look at post #4. Can you do the algebra?
You said that you already know the answer: $\displaystyle \dfrac{-1}{(x+1)^2}$
So work towards that answer.

7. Got it thanks!